CHAPTER 4 IMPERFECTIONS IN SOLIDS PROBLEM SOLUTIONS


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1 41 CHAPTER 4 IMPERFECTIONS IN SOLIDS PROBLEM SOLUTIONS Vacancies and SelfInterstitials 4.1 In order to compute the fraction of atom sites that are vacant in copper at 1357 K, we must employ Equation 4.1. As stated in the problem, Q v 0.90 ev/atom. Thus, N v N exp Q v exp kt 0.90 ev /atom ( ev/atom K) (1357 K) 4.56 x 104
2 Determination of the number of vacancies per cubic meter in gold at 900 C (1173 K) requires the utilization of Equations 4.1 and 4.2 as follows: N v N exp Q v N A ρ Au kt A Au exp Q v kt ( atoms /mol)(18.63 g /cm 3 ) g /mol exp 0.98 ev /atom ( ev/atom K) (1173 K) 3.52 x cm x m 3
3 This problem calls for the computation of the activation energy for vacancy formation in silver. Upon examination of Equation 4.1, all parameters besides Q v are given except N, the total number of atomic sites. However, N is related to the density, (ρ), Avogadro's number (N A ), and the atomic weight (A) according to Equation 4.2 as N N A ρ Pb A Pb ( atoms / mol)(9.5 g /cm 3 ) g /mol 5.30 x atoms/cm x atoms/m 3 Now, taking natural logarithms of both sides of Equation 4.1, ln N v lnn Q v kt and, after some algebraic manipulation N Q v kt ln v N ( ev/atom K)(800 C K) ln m m ev/atom
4 44 Impurities in Solids 4.4 In this problem we are asked to cite which of the elements listed form with Ni the three possible solid solution types. For complete substitutional solubility the following criteria must be met: 1) the difference in atomic radii between Ni and the other element ( R%) must be less than ±15%, 2) the crystal structures must be the same, 3) the electronegativities must be similar, and 4) the valences should be the same, or nearly the same. Below are tabulated, for the various elements, these criteria. Crystal Electro Element R% Structure negativity Valence Ni FCC 2+ C 43 H 63 O 52 Ag +16 FCC Al +15 FCC Co +0.6 HCP 0 2+ Cr +0.2 BCC Fe 0.4 BCC 0 2+ Pt +11 FCC Zn +7 HCP (a) Pt is the only element that meets all of the criteria and thus forms a substitutional solid solution having complete solubility. At elevated temperatures Co and Fe experience allotropic transformations to the FCC crystal structure, and thus display complete solid solubility at these temperatures. (b) Ag, Al, Co, Cr, Fe, and Zn form substitutional solid solutions of incomplete solubility. All these metals have either BCC or HCP crystal structures, and/or the difference between their atomic radii and that for Ni are greater than ±15%, and/or have a valence different than 2+. (c) C, H, and O form interstitial solid solutions. These elements have atomic radii that are significantly smaller than the atomic radius of Ni.
5 In the drawing below is shown the atoms on the () face of an FCC unit cell; the interstitial site is at the center of the edge. The diameter of an atom that will just fit into this site (2r) is just the difference between that unit cell edge length (a) and the radii of the two host atoms that are located on either side of the site (R); that is 2r a 2R However, for FCC a is related to R according to Equation 3.1 as a 2R 2 ; therefore, solving for r from the above equation gives r a 2 R 2 A () face of a BCC unit cell is shown below. 2 R 2 2 R R
6 46 The interstitial atom that just fits into this interstitial site is shown by the small circle. It is situated in the plane of this () face, midway between the two vertical unit cell edges, and one quarter of the distance between the bottom and top cell edges. From the right triangle that is defined by the three arrows we may write a a 2 4 (R + r) 2 However, from Equation 3.3, a 4R, and, therefore, making this substitution, the above equation takes the form 3 4R R R 2 +2Rr + r 2 After rearrangement the following quadratic equation results: r 2 + 2Rr 0.667R 2 0 And upon solving for r: r (2R) ± (2R)2 (4)(1)( 0.667R 2 ) 2
7 47 2R ± 2.582R 2 And, finally r(+) r( ) 2R R R 2R 2.582R R Of course, only the r(+) root is possible, and, therefore, r 0.291R. BCC. Thus, for a host atom of radius R, the size of an interstitial site for FCC is approximately 1.4 times that for
8 48 Specification of Composition 4.6 (a) This problem asks that we derive Equation 4.7a. To begin, C 1 is defined according to Equation 4.3 as C 1 m 1 m 1 + m 2 or, equivalently C 1 m 1 ' m 1 ' + m 2 ' where the primed m's indicate masses in grams. From Equation 4.4 we may write m 1 ' n m1 A 1 m 2 ' n m2 A 2 And, substitution into the C 1 expression above C 1 n m1 A 1 n m1 A 1 + n m2 A 2 From Equation 4.5 it is the case that n m1 C 1 ' (n m1 + n m2 ) n m2 C 2 ' (n m1 + n m2 ) And substitution of these expressions into the above equation leads to
9 49 C 1 C 1 ' A 1 C 1 ' A 1 + C 2 ' A 2 which is just Equation 4.7a. (b) This problem asks that we derive Equation 4.9a. To begin, C " 1 is defined as the mass of component 1 per unit volume of alloy, or C 1 " m 1 V If we assume that the total alloy volume V is equal to the sum of the volumes of the two constituentsi.e., V V 1 + V 2 then C 1 " m 1 V 1 + V 2 Furthermore, the volume of each constituent is related to its density and mass as V 1 m 1 ρ 1 V 2 m 2 ρ 2 This leads to C 1 " m 1 m 1 + m 2 ρ 1 ρ 2 From Equation 4.3, m 1 and m 2 may be expressed as follows: m 1 C 1 (m 1 + m 2 )
10 410 m 2 C 2 (m 1 + m 2 ) Substitution of these equations into the preceding expression yields C 1 (m 1 + m 2 ) C" 1 C 1 (m 1 + m 2 ) C 2 (m 1 + m 2 ) ρ 1 + ρ 2 C 1 C 1 + C 2 ρ 1 ρ 2 If the densities ρ 1 and ρ 2 are given in units of g/cm 3, then conversion to units of kg/m 3 requires that we multiply this equation by 10 3, inasmuch as 1 g/cm kg/m 3 Therefore, the previous equation takes the form C 1 " C 1 C 1 + C 2 ρ 1 ρ 2 x 10 3 which is the desired expression. (c) Now we are asked to derive Equation 4.10a. The density of an alloy ρ ave is just the total alloy mass M divided by its volume V ρ ave M V Or, in terms of the component elements 1 and 2
11 411 ρ ave m 1 + m 2 V 1 + V 2 [Note: here it is assumed that the total alloy volume is equal to the separate volumes of the individual components, which is only an approximation; normally V will not be exactly equal to (V 1 + V 2 )]. Each of V 1 and V 2 may be expressed in terms of its mass and density as, V 1 m 1 ρ 1 V 2 m 2 ρ 2 When these expressions are substituted into the above equation, we get ρ ave m 1 + m 2 m 1 ρ 1 + m 2 ρ 2 Furthermore, from Equation 4.3 m 1 C 1 (m 1 + m 2 ) m 2 C 2 (m 1 + m 2 ) Which, when substituted into the above ρ ave expression yields ρ ave m 1 + m 2 C 1 (m 1 + m 2 ) C 2 (m 1 + m 2 ) ρ 1 + ρ 2 And, finally, this equation reduces to C 1 + C 2 ρ 1 ρ 2
12 413 Equation 4.6 as 4.7 In order to compute composition, in atom percent, of a 92.5 wt% Ag7.5 wt% Cu alloy, we employ C ' Ag C Ag A Cu C Ag A Cu + C Cu A Ag (92.5)(63.55 g /mol) (92.5)(63.55 g /mol) + (7.5)(107.87g /mol) 87.9 at% C ' Cu C Cu A Ag C Ag A Cu + C Cu A Ag (7.5)( g /mol) (92.5)(63.55 g /mol) + (7.5)(107.87g /mol) 12.1 at%
13 414 Equation 4.7 as 4.8 In order to compute composition, in weight percent, of a 5 at% Cu95 at% Pt alloy, we employ C Cu C ' Cu A Cu C ' Cu A Cu + C ' Pt A Pt (5)(63.55 g /mol) (5)(63.55 g / mol) + (95)( g /mol) 1.68 wt% C Pt C ' Pt A Pt C ' Cu A Cu + C ' Pt A Pt (95)( g /mol) (5)(63.55 g /mol) + (95)( g /mol) x wt%
14 The concentration, in weight percent, of an element in an alloy may be computed using a modified form of Equation 4.3. For this alloy, the concentration of iron (C Fe ) is just C Fe m Fe m Fe + m C + m Cr 105 kg 105 kg kg kg wt% Similarly, for carbon C C 0.2 kg 105 kg kg kg 0.19 wt% And for chromium C Cr 1.0 kg 105 kg kg kg 0.94 wt%
15 The concentration of an element in an alloy, in atom percent, may be computed using Equation 4.5. However, it first becomes necessary to compute the number of moles of both Cu and Zn, using Equation 4.4. Thus, the number of moles of Cu is just n mcu m ' Cu A Cu 33 g g /mol mol Likewise, for Zn n mzn 47 g g /mol mol Now, use of Equation 4.5 yields C ' Cu n mcu n mcu + n mzn mol mol mol 41.9 at% Also, C ' Zn mol mol mol 58.1 at%
16 In this problem we are asked to determine the concentrations, in atom percent, of the AgAuCu alloy. It is first necessary to convert the amounts of Ag, Au, and Cu into grams. m ' Ag (44.5 lb m )(453.6 g/lb m ) 20,185 g m ' Au (83.7 lb m )(453.6 g/lb m ) 37,966 g m ' Cu (5.3 lb m )(453.6 g/lb m ) 2, 404 g These masses must next be converted into moles (Equation 4.4), as n mag m ' Ag A Ag 20,185 g g /mol mol n mau 37,966 g g / mol mol n mcu 2, 404 g g / mol 37.8 mol Now, employment of a modified form of Equation 4.5, gives C ' Ag n mag n mag + n mau + n mcu mol mol mol mol 44.8 at% C ' Au mol mol mol mol 46.2 at% C ' Cu 37.8 mol mol mol mol 9.0 at%
17 leads to 4.12 We are asked to compute the composition of a PbSn alloy in atom percent. Employment of Equation C ' Pb C Pb A Sn C Pb A Sn + C Sn A Pb 5.5( g / mol) 5.5( g / mol) (207.2 g / mol) 3.2 at% C ' Sn C Sn A Pb C Sn A Pb + C Pb A Sn 94.5(207.2 g / mol) 94.5(207.2 g /mol) + 5.5( g /mol) 96.8 at%
18 This problem calls for a conversion of composition in atom percent to composition in weight percent. The composition in atom percent for Problem 4.11 is 44.8 at% Ag, 46.2 at% Au, and 9.0 at% Cu. Modification of Equation 4.7 to take into account a threecomponent alloy leads to the following C Ag C ' Ag C ' Ag A Ag A Ag + C ' Au A Au + C ' Cu A Cu (44.8) ( g /mol) (44.8)( g /mol) + (46.2) ( g /mol) + (9.0) (63.55 g /mol) 33.3 wt% C Au C ' Ag C ' Au A Au A Ag + C ' Au A Au + C ' Cu A Cu (46.2) ( g /mol) (44.8)( g /mol) + (46.2) ( g /mol) + (9.0) (63.55 g /mol) 62.7 wt% C Cu C ' Ag C ' Cu A Cu A Ag + C ' Au A Au + C ' Cu A Cu (9.0) (63.55 g /mol) (44.8)( g /mol) + (46.2) ( g /mol) + (9.0) (63.55 g /mol) 4.0 wt%
19 This problem calls for a determination of the number of atoms per cubic meter for lead. In order to solve this problem, one must employ Equation 4.2, N N A ρ Pb A Pb The density of Pb (from the table inside of the front cover) is g/cm 3, while its atomic weight is g/mol. Thus, N (6.023 x 1023 atoms /mol)(11.35 g /cm 3 ) g /mol 3.30 x atoms/cm x atoms/m 3
20 In order to compute the concentration in kg/m 3 of Si in a 0.25 wt% Si wt% Fe alloy we must employ Equation 4.9 as C " C Si Si C Si + C Fe ρ Si ρ Fe 10 3 From inside the front cover, densities for silicon and iron are 2.33 and 7.87 g/cm 3, respectively; and, therefore C " Si g /cm g /cm kg/m 3
21 We are asked in this problem to determine the approximate density of a Ti6Al4V titanium alloy that has a composition of 90 wt% Ti, 6 wt% Al, and 4 wt% V. In order to solve this problem, Equation 4.10a is modified to take the following form: ρ ave C Ti ρ Ti + C Al ρ Al + C V ρ V And, using the density values for Ti, Al, and V i.e., 4.51 g/cm 3, 2.71 g/cm 3, and 6.10 g/cm 3 (as taken from inside the front cover of the text), the density is computed as follows: ρ ave 90 wt% 4.51 g /cm wt% 2.71 g /cm wt% 6.10 g /cm g/cm 3
22 This problem asks that we determine the unit cell edge length for a 80 wt% Ag20 wt% Pd alloy. In order to solve this problem it is necessary to employ Equation 3.5; in this expression density and atomic weight will be averages for the alloy that is ρ ave na ave V C N A Inasmuch as the unit cell is cubic, then V C a 3, then ρ ave na ave a 3 N A And solving this equation for the unit cell edge length, leads to a na 1/3 ave ρ ave N A Expressions for A ave and ρ ave are found in Equations 4.11a and 4.10a, respectively, which, when incorporated into the above expression yields a n C Ag C Ag + C Pd A Ag A Pd ρ Ag + C Pd ρ Pd 1/3 N A Since the crystal structure is FCC, the value of n in the above expression is 4 atoms per unit cell. The atomic weights for Ag and Pd are and g/mol, respectively (Figure 2.6), whereas the densities for the Ag and Pd are g/cm 3 (inside front cover) and g/cm 3. Substitution of these, as well as the concentration values stipulated in the problem statement, into the above equation gives
23 424 1/3 (4 atoms/unit cell) 80 wt% g/mol + 20 wt% g/mol a atoms/mol 80 wt% g/cm wt% ( ) g/cm cm nm
24 This problem asks that we determine, for a hypothetical alloy that is composed of 25 wt% of metal A and 75 wt% of metal B, whether the crystal structure is simple cubic, facecentered cubic, or bodycentered cubic. We are given the densities of the these metals (ρ A 6.17 g/cm 3 and ρ B 8.00 g/cm 3 for B), their atomic weights (A A g/mol and A B g/mol), and that the unit cell edge length is nm (i.e., 3.32 x 108 cm). In order to solve this problem it is necessary to employ Equation 3.5; in this expression density and atomic weight will be averages for the alloy that is ρ ave na ave V C N A Inasmuch as for each of the possible crystal structures, the unit cell is cubic, then V C a 3, or ρ ave na ave a 3 N A And, in order to determine the crystal structure it is necessary to solve for n, the number of atoms per unit cell. For n 1, the crystal structure is simple cubic, whereas for n values of 2 and 4, the crystal structure will be either BCC or FCC, respectively. When we solve the above expression for n the result is as follows: n ρ ave a3 N A A ave Expressions for A ave and ρ ave are found in Equations 4.11a and 4.10a, respectively, which, when incorporated into the above expression yields n C A + C a B 3 N A ρ A ρ B C A + C B A A A B Substitution of the concentration values (i.e., C A 25 wt% and C B 75 wt%) as well as values for the other parameters given in the problem statement, into the above equation gives
25 426 n ( nm) 3 ( atoms/mol) 25 wt% 6.17 g/cm wt% 8.00 g/cm 3 25 wt% g/mol + 75 wt% g/mol 1.00 atom/unit cell Therefore, on the basis of this value, the crystal structure is simple cubic.
26 This problem asks that we derive Equation 4.18, using other equations given in the chapter. The concentration of component 1 in atom percent (C ' 1 ) is just c ' 1 where c ' 1 is the atom fraction of component 1. Furthermore, c ' 1 is defined as c ' 1 N 1 /N where N 1 and N are, respectively, the number of atoms of component 1 and total number of atoms per cubic centimeter. Thus, from the above discussion the following holds: N 1 C 1 ' N Substitution into this expression of the appropriate form of N from Equation 4.2 yields N 1 C 1 ' N A ρ ave A ave And, finally, substitution into this equation expressions for C ' 1 (Equation 4.6a), ρ ave (Equation 4.10a), A ave (Equation 4.11a), and realizing that C 2 (C 1 ), and after some algebraic manipulation we obtain the desired expression: N 1 N A C 1 C 1 A 1 + A 1 ( C ρ 1 ρ 1 ) 2
27 This problem asks us to determine the number of molybdenum atoms per cubic centimeter for a 16.4 wt% Mo83.6 wt% W solid solution. To solve this problem, employment of Equation 4.18 is necessary, using the following values: C 1 C Mo 16.4 wt% ρ 1 ρ Mo g/cm 3 ρ 2 ρ W 19.3 g/cm 3 A 1 A Mo g/mol Thus N Mo C Mo A Mo ρ Mo N A C Mo + A Mo ρ W ( C Mo ) ( atoms /mol) (16.4 wt%) (16.4 wt%)(95.94 g /mol) g /mol g /cm 3 + ( 16.4 wt%) 19.3 g /cm x atoms/cm 3
28 This problem asks us to determine the number of niobium atoms per cubic centimeter for a 24 wt% Nb76 wt% V solid solution. To solve this problem, employment of Equation 4.18 is necessary, using the following values: C 1 C Nb 24 wt% ρ 1 ρ Nb 8.57 g/cm 3 ρ 2 ρ V 6.10 g/cm 3 A 1 A Nb g/mol Thus N Nb C Nb A Nb ρ Nb N A C Nb + A Nb ρ V ( C Nb ) ( atoms /mol) (24 wt%) (24 wt%)(92.91 g /mol) g /mol 8.57 g /cm 3 + ( 24 wt%) 6.10 g /cm x atoms/cm 3
29 This problem asks that we derive Equation 4.19, using other equations given in the chapter. The number of atoms of component 1 per cubic centimeter is just equal to the atom fraction of component 1 (c ' 1 ) times the total number of atoms per cubic centimeter in the alloy (N). Thus, using the equivalent of Equation 4.2, we may write N 1 c 1 ' N c 1 ' N A ρ ave A ave Realizing that c ' 1 C 1 ' and C 2 ' C 1 ' and substitution of the expressions for ρ ave and A ave, Equations 4.10b and 4.11b, respectively, leads to N 1 c 1 ' N A ρ ave A ave N A C 1 ' ρ 1 ρ 2 C 1 ' ρ 2 A 1 + ( C 1 ' )ρ 1 A 2 And, solving for C 1 ' C 1 ' N 1 ρ 1 A 2 N A ρ 1 ρ 2 N 1 ρ 2 A 1 + N 1 ρ 1 A 2 Substitution of this expression for C ' 1 into Equation 4.7a, which may be written in the following form
30 431 C 1 C 1 ' A 1 C 1 ' A 1 + C 2 ' A 2 C 1 ' A 1 C 1 ' A 1 + ( C 1 ' )A 2 yields C N A ρ 2 N 1 A 1 ρ 2 ρ 1 the desired expression.
31 This problem asks us to determine the weight percent of Au that must be added to Ag such that the resultant alloy will contain 5.5 x Au atoms per cubic centimeter. To solve this problem, employment of Equation 4.19 is necessary, using the following values: N 1 N Au 5.5 x atoms/cm 3 ρ 1 ρ Au g/cm 3 ρ 2 ρ Ag g/cm 3 A 1 A Au g/mol A 2 A Ag g/mol Thus C Au 1 + N A ρ Ag N Au A Au ρ Ag ρ Au 1 + ( atoms /mol)(10.49 g /cm 3 ) ( atoms /cm 3 )( g /mol) g /cm g /cm wt%
32 This problem asks us to determine the weight percent of Ge that must be added to Si such that the resultant alloy will contain 2.43 x10 21 Ge atoms per cubic centimeter. To solve this problem, employment of Equation 4.19 is necessary, using the following values: N 1 N Ge 2.43 x atoms/cm 3 ρ 1 ρ Ge 5.32 g/cm 3 ρ 2 ρ Si 2.33 g/cm 3 A 1 A Ge g/mol A 2 A Si g/mol Thus C Ge 1 + N A ρ Si ρ Si N Ge A Ge ρ Ge 1 + ( atoms /mol)(2.33 g /cm 3 ) ( atoms /cm 3 ) (72.64 g / mol) 2.33 g /cm g /cm wt%
33 This problems asks that we compute the unit cell edge length for a 90 wt% Fe10 wt% V alloy. First of all, the atomic radii for Fe and V (using the table inside the front cover) are and nm, respectively. Also, using Equation 3.5 it is possible to compute the unit cell volume, and inasmuch as the unit cell is cubic, the unit cell edge length is just the cube root of the volume. However, it is first necessary to calculate the density and average atomic weight of this alloy using Equations 4.10a and 4.11a. Inasmuch as the densities of iron and vanadium are 7.87g/cm 3 and 6.10 g/cm 3, respectively, (as taken from inside the front cover), the average density is just ρ ave C V ρ V + C Fe ρ Fe 10 wt% 6.10 g /cm wt% 7.87 g /cm g/cm 3 And for the average atomic weight A ave C V + C Fe A V A Fe 10 wt% g /mole + 90 wt% g /mol g/mol Now, V C is determined from Equation 3.5 as V C na ave ρ ave N A (2 atoms /unit cell)(55.32 g /mol) (7.65 g /cm 3 )( atoms /mol)
34 x cm 3 /unit cell And, finally a (V C ) 1/3 ( cm 3 /unit cell) 1/ x 108 cm nm
35 436 Dislocations Linear Defects 4.26 The Burgers vector and dislocation line are perpendicular for edge dislocations, parallel for screw dislocations, and neither perpendicular nor parallel for mixed dislocations.
36 437 Interfacial Defects 4.27 The surface energy for a crystallographic plane will depend on its packing density [i.e., the planar density (Section 3.11)] that is, the higher the packing density, the greater the number of nearestneighbor atoms, and the more atomic bonds in that plane that are satisfied, and, consequently, the lower the surface energy. From 1 the solution to Problem 3.53, planar densities for FCC () and (111) planes are 4R 2 and 1 2R 2 that is and R2 R 2 the lower surface energy. 3, respectively (where R is the atomic radius). Thus, since the planar density for (111) is greater, it will have
37 The surface energy for a crystallographic plane will depend on its packing density [i.e., the planar density (Section 3.11)] that is, the higher the packing density, the greater the number of nearestneighbor atoms, and the more atomic bonds in that plane that are satisfied, and, consequently, the lower the surface energy. From 3 the solution to Problem 3.54, the planar densities for BCC () and (110) are 16R 2 and 3 8R 2 2, respectively that is and. Thus, since the planar density for (110) is greater, it will have the lower surface energy. R2 R2
38 (a) The surface energy will be greater than the grain boundary energy. For grain boundaries, some atoms on one side of a boundary will bond to atoms on the other side; such is not the case for surface atoms. Therefore, there will be fewer unsatisfied bonds along a grain boundary. (b) The smallangle grain boundary energy is lower than for a highangle one because more atoms bond across the boundary for the smallangle, and, thus, there are fewer unsatisfied bonds.
39 (a) A twin boundary is an interface such that atoms on one side are located at mirror image positions of those atoms situated on the other boundary side. The region on one side of this boundary is called a twin. (b) Mechanical twins are produced as a result of mechanical deformation and generally occur in BCC and HCP metals. Annealing twins form during annealing heat treatments, most often in FCC metals.
40 (a) The interfacial defect that exists for this stacking sequence is a twin boundary, which occurs at the indicated position. The stacking sequence on one side of this position is mirrored on the other side. (b) The interfacial defect that exists within this FCC stacking sequence is a stacking fault, which occurs between the two lines. Within this region, the stacking sequence is HCP.
41 442 Grain Size Determination 4.32 (a) This problem calls for a determination of the average grain size of the specimen which microstructure is shown in Figure 4.14(b). Seven line segments were drawn across the micrograph, each of which was 60 mm long. The average number of grain boundary intersections for these lines was 8.7. Therefore, the average line length intersected is just 60 mm mm Hence, the average grain diameter, d, is d ave. line length intersected magnification 6.9 mm mm (b) This portion of the problem calls for us to estimate the ASTM grain size number for this same material. The average grain size number, n, is related to the number of grains per square inch, N, at a magnification of x according to Equation Inasmuch as the magnification is x, the value of N is measured directly from the micrograph, which is approximately 12 grains. In order to solve for n in Equation 4.16, it is first necessary to take logarithms as log N (n 1) log 2 From which n equals n log N log log 12 log
42 (a) This portion of the problem calls for a determination of the average grain size of the specimen which microstructure is shown in Figure 9.25(a). Seven line segments were drawn across the micrograph, each of which was 60 mm long. The average number of grain boundary intersections for these lines was 6.3. Therefore, the average line length intersected is just 60 mm mm Hence, the average grain diameter, d, is d ave. line length intersected magnification 9.5 mm mm (b) This portion of the problem calls for us to estimate the ASTM grain size number for this same material. The average grain size number, n, is related to the number of grains per square inch, N, at a magnification of x according to Equation However, the magnification of this micrograph is not x, but rather 90x. Consequently, it is necessary to use Equation 4.17 M 2 N M 2 n 1 where N M the number of grains per square inch at magnification M, and n is the ASTM grain size number. Taking logarithms of both sides of this equation leads to the following: log N M M + 2 log (n 1) log 2 Solving this expression for n gives n M log N M + 2 log + 1 log 2 From Figure 9.25(a), N M is measured to be approximately 4, which leads to n log log log 2 2.7
43 (a) This part of problem asks that we compute the number of grains per square inch for an ASTM grain size of 6 at a magnification of x. All we need do is solve for the parameter N in Equation 4.16, inasmuch as n 6. Thus N 2 n grains/in. 2 (b) Now it is necessary to compute the value of N for no magnification. In order to solve this problem it is necessary to use Equation 4.17: M 2 N M 2 n 1 where N M the number of grains per square inch at magnification M, and n is the ASTM grain size number. Without any magnification, M in the above equation is 1, and therefore, 1 2 N And, solving for N 1, N 1 320,000 grains/in. 2.
44 This problem asks that we determine the ASTM grain size number if 30 grains per square inch are measured at a magnification of 250. In order to solve this problem we make use of Equation 4.17: 2 M N M 2 n 1 where N M the number of grains per square inch at magnification M, and n is the ASTM grain size number. Solving the above equation for n, and realizing that N M 30, while M 250, we have n M log N M + 2 log + 1 log 2 log log 250 log
45 This problem asks that we determine the ASTM grain size number if 25 grains per square inch are measured at a magnification of 75. In order to solve this problem we make use of Equation 4.17 viz. M 2 N M 2 n 1 where N M the number of grains per square inch at magnification M, and n is the ASTM grain size number. Solving the above equation for n, and realizing that N M 25, while M 75, we have n M log N M + 2 log + 1 log 2 75 log log log
46 447 DESIGN PROBLEMS Specification of Composition 4.D1 This problem calls for us to compute the concentration of lithium (in wt%) that, when added to aluminum, will yield a density of 2.47 g/cm 3. Solution of this problem requires the use of Equation 4.10a, which takes the form ρ ave C Li ρ Li + C Li ρ Al inasmuch as C Li + C Al. According to the table inside the front cover, the respective densities of Li and Al are and 2.71 g/cm 3. Upon solving for C Li from the above equation, we get C Li ρ Li (ρ Al ρ ave ) ρ ave (ρ Al ρ Li ) ()(0.534 g /cm3 )(2.71 g /cm g /cm 3 ) (2.47 g /cm 3 )(2.71 g /cm g /cm 3 ) 2.38 wt%
47 D2 This problem asks that we determine the concentration (in weight percent) of Cu that must be added to Pt so as to yield a unit cell edge length of nm. To begin, it is necessary to employ Equation 3.5, and solve for the unit cell volume, V C, as V C na ave ρ ave N A where A ave and ρ ave are the atomic weight and density, respectively, of the PtCu alloy. Inasmuch as both of these materials have the FCC crystal structure, which has cubic symmetry, V C is just the cube of the unit cell length, a. That is V C a 3 (0.390 nm) 3 ( cm) cm 3 It is now necessary to construct expressions for A ave and ρ ave in terms of the concentration of vanadium, C Cu, using Equations 4.11a and 4.10a. For A ave we have A ave C Cu + ( C Cu ) A Cu A Pt C Cu g /mol + ( C Cu ) g /mol whereas for ρ ave ρ ave C Cu + ( C Cu ) ρ Cu ρ Pt C Cu 8.94 g /cm 3 + ( C Cu ) g /cm 3
48 449 Within the FCC unit cell there are 4 equivalent atoms, and thus, the value of n in Equation 3.5 is 4; hence, this expression may be written in terms of the concentration of Cu in weight percent as follows: V C x cm 3 na ave ρ ave N A (4 atoms /unit cell) C Cu g /mol + ( C Cu ) g /mol C Cu 8.94 g /cm 3 + ( C Cu ) ( atoms /mol) g /cm 3 And solving this expression for C Cu leads to C Cu wt%.
v kt = N A ρ Au exp (
42 4.2 Determination of the number of vacancies per cubic meter in gold at 900 C (1173 K) requires the utilization of Equations 4.1 and 4.2 as follows: N v N exp Q v N A ρ Au kt A Au exp Q v kt (6.023
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